Find the Inverse (7x-3)/16

$\frac{7 x - 3}{16}$

Step 1

Interchange the variables.

$x = \frac{7 y - 3}{16}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{7 y - 3}{16} = x$ .

$\frac{7 y - 3}{16} = x$

Multiply both sides by $16$ .

$\frac{7 y - 3}{16} \cdot 16 = x \cdot 16$

Simplify.

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{7 y - 3}{16} \cdot 16 = x \cdot 16$

Rewrite the expression.

$7 y - 3 = x \cdot 16$

Simplify the right side.

Move $16$ to the left of $x$ .

$7 y - 3 = 16 x$

Solve for $y$ .

Add $3$ to both sides of the equation.

$7 y = 16 x + 3$

Divide each term in $7 y = 16 x + 3$ by $7$ and simplify.

Divide each term in $7 y = 16 x + 3$ by $7$ .

$\frac{7 y}{7} = \frac{16 x}{7} + \frac{3}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 y}{7} = \frac{16 x}{7} + \frac{3}{7}$

Divide $y$ by $1$ .

$y = \frac{16 x}{7} + \frac{3}{7}$

Step 3

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \frac{16 x}{7} + \frac{3}{7}$

Step 4

Verify if $f^{- 1} (x) = \frac{16 x}{7} + \frac{3}{7}$ is the inverse of $f (x) = \frac{7 x - 3}{16}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\frac{7 x - 3}{16})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{7 x - 3}{16}) = \frac{16 (\frac{7 x - 3}{16})}{7} + \frac{3}{7}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{7 x - 3}{16}) = \frac{16 (\frac{7 x - 3}{16}) + 3}{7}$

Cancel the common factor of $16$ .

Cancel the common factor.

$f^{- 1} (\frac{7 x - 3}{16}) = \frac{16 (\frac{7 x - 3}{16}) + 3}{7}$

Rewrite the expression.

$f^{- 1} (\frac{7 x - 3}{16}) = \frac{7 x - 3 + 3}{7}$

Combine the opposite terms in $7 x - 3 + 3$ .

Add $- 3$ and $3$ .

$f^{- 1} (\frac{7 x - 3}{16}) = \frac{7 x + 0}{7}$

Add $7 x$ and $0$ .

$f^{- 1} (\frac{7 x - 3}{16}) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} (\frac{7 x - 3}{16}) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f^{- 1} (\frac{7 x - 3}{16}) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\frac{16 x}{7} + \frac{3}{7})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{16 x}{7} + \frac{3}{7}) = \frac{7 (\frac{16 x}{7} + \frac{3}{7}) - 3}{16}$

Simplify the numerator.

Apply the distributive property.

$f (\frac{16 x}{7} + \frac{3}{7}) = \frac{7 (\frac{16 x}{7}) + 7 (\frac{3}{7}) - 3}{16}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (\frac{16 x}{7} + \frac{3}{7}) = \frac{7 (\frac{16 x}{7}) + 7 (\frac{3}{7}) - 3}{16}$

Rewrite the expression.

$f (\frac{16 x}{7} + \frac{3}{7}) = \frac{16 x + 7 (\frac{3}{7}) - 3}{16}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (\frac{16 x}{7} + \frac{3}{7}) = \frac{16 x + 7 (\frac{3}{7}) - 3}{16}$

Rewrite the expression.

$f (\frac{16 x}{7} + \frac{3}{7}) = \frac{16 x + 3 - 3}{16}$

Subtract $3$ from $3$ .

$f (\frac{16 x}{7} + \frac{3}{7}) = \frac{16 x + 0}{16}$

Add $16 x$ and $0$ .

$f (\frac{16 x}{7} + \frac{3}{7}) = \frac{16 x}{16}$

Cancel the common factor of $16$ .

Cancel the common factor.

$f (\frac{16 x}{7} + \frac{3}{7}) = \frac{16 x}{16}$

Divide $x$ by $1$ .

$f (\frac{16 x}{7} + \frac{3}{7}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{16 x}{7} + \frac{3}{7}$ is the inverse of $f (x) = \frac{7 x - 3}{16}$ .

$f^{- 1} (x) = \frac{16 x}{7} + \frac{3}{7}$

Find the Inverse (7x-3)/16

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